Jika jumlah n suku pertama sebuah deret dinyatakanoleh Sn = 3n^2 - 4n, maka rumus suku ke-n adalah:
a. Un = 3n - 4
b. Un = 6n - 1
c. Un = 6n + 4
d. Un = 3n - 1
e. Un = 4n - 3
a. Un = 3n - 4
b. Un = 6n - 1
c. Un = 6n + 4
d. Un = 3n - 1
e. Un = 4n - 3
izharulhaq78
![\text{penjelasan :}\\\\U_n=S_n-S_{n-1}\\U_n=pn^2+qn-(p[n-1]^2+q[n-1])\\U_n=pn^2+qn-p[n^2-2n+1]-q[n-1]\\U_n=pn^2-pn^2+2pn-p+qn-qn+q\\U_n=2pn-p+q\\U_n=p(2n-1)+q](https://tex.z-dn.net/?f=%5Ctext%7Bpenjelasan+%3A%7D%5C%5C%5C%5CU_n%3DS_n-S_%7Bn-1%7D%5C%5CU_n%3Dpn%5E2%2Bqn-%28p%5Bn-1%5D%5E2%2Bq%5Bn-1%5D%29%5C%5CU_n%3Dpn%5E2%2Bqn-p%5Bn%5E2-2n%2B1%5D-q%5Bn-1%5D%5C%5CU_n%3Dpn%5E2-pn%5E2%2B2pn-p%2Bqn-qn%2Bq%5C%5CU_n%3D2pn-p%2Bq%5C%5CU_n%3Dp%282n-1%29%2Bq "\text{penjelasan :}\\\\U_n=S_n-S_{n-1}\\U_n=pn^2+qn-(p[n-1]^2+q[n-1])\\U_n=pn^2+qn-p[n^2-2n+1]-q[n-1]\\U_n=pn^2-pn^2+2pn-p+qn-qn+q\\U_n=2pn-p+q\\U_n=p(2n-1)+q")
1 votes
Thanks 1
tiaratfe
maaf, 2n-1 darimana ya? bisa tolong jelaskan?
izharulhaq78
sudah ditambahkan penjelasan agar lebih jelas
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